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Vandermonde conditioning improves after centering and scaling the nodes

Example

Interpolating a linear polynomial through two points whose nodes are $t_{0} = 100$ and $t_{1} = 101$ is the Vandermonde system $V (t_{0}, t_{1}) c = y$ with $V (t_{0}, t_{1}) = (\begin{array}{cc} 1 & t_{0} \\ 1 & t_{1} \end{array})$ , which has $κ_{\infty} = 20502$ . Centering and scaling the nodes to $s_{0} = - 1 / 2$ and $s_{1} = 1 / 2$ through the affine change of variable $t = s + 201 / 2$ gives the equivalent system for the same interpolation task with $κ_{\infty} = 3$ . Conditioning is a property of the chosen coordinates, not of the underlying interpolation problem.

Facts & Assumptions

Given: The nodes $t_{0} = 100$ , $t_{1} = 101$ and their images $s_{0} = - 1 / 2$ , $s_{1} = 1 / 2$ under $t = s + 201 / 2$ , and the Vandermonde matrices $V (t_{0}, t_{1}) = (\begin{array}{cc} 1 & t_{0} \\ 1 & t_{1} \end{array})$ , $V (s_{0}, s_{1}) = (\begin{array}{cc} 1 & s_{0} \\ 1 & s_{1} \end{array})$ .

[L1]

$κ_{p} (A) = ∥ A ∥_{p} ∥ A^{- 1} ∥_{p}$ (The condition number kappa_p(A) = ||A||_p ||A^{-1}||_p of a nonsingular linear system).

[L2]

The induced $\infty$ -norm is the maximum row sum (The induced 1-norm is the maximum column sum and the induced infinity-norm is the maximum row sum).

[L3]

For a nonsingular $2 \times 2$ matrix $M = (\begin{array}{cc} a & b \\ c & d \end{array})$ , $M^{- 1} = (a d - b c)^{- 1} (\begin{array}{cc} d & - b \\ - c & a \end{array})$ .

Verification

technique · direct

1.1

For the node pair $(t_{0}, t_{1}) = (100, 101)$ , the determinant is $101 - 100 = 1$ , and [L3] gives $V (t_{0}, t_{1})^{- 1} = (\begin{array}{cc} 101 & - 100 \\ - 1 & 1 \end{array})$ .

L3algebra

1.2

By [L2], $∥ V (100, 101) ∥_{\infty} = \max {101, 102} = 102$ and $∥ V (100, 101)^{- 1} ∥_{\infty} = \max {201, 2} = 201$ , so [L1] gives $κ_{\infty} = 102 \cdot 201 = 20502$ .

L1L2algebra

1.3

For the centered and scaled pair $(s_{0}, s_{1}) = (- 1 / 2, 1 / 2)$ , the determinant is $1 / 2 - (- 1 / 2) = 1$ , and [L3] gives $V (s_{0}, s_{1})^{- 1} = (\begin{array}{cc} 1 / 2 & 1 / 2 \\ - 1 & 1 \end{array})$ .

L3algebra

1.4

By [L2], $∥ V (s_{0}, s_{1}) ∥_{\infty} = \max {3 / 2, 3 / 2} = 3 / 2$ and $∥ V (s_{0}, s_{1})^{- 1} ∥_{\infty} = \max {1, 2} = 2$ , so [L1] gives $κ_{\infty} = (3 / 2) \cdot 2 = 3$ .

L1L2algebra

2.1

The change of variable $t = s + 201 / 2$ maps $s_{0}$ to $100$ and $s_{1}$ to $101$ , and interpolating the same data $(t_{i}, y_{i})$ in the coordinate $t$ is the system $V (t_{0}, t_{1}) c = y$ while interpolating it in the coordinate $s$ is $V (s_{0}, s_{1}) c^{'} = y$ with $c_{0}^{'} = c_{0} + (201 / 2) c_{1}$ and $c_{1}^{'} = c_{1}$ ; the two systems represent the same polynomial task, yet their condition numbers are $20502$ and $3$ .

step 1.2step 1.4algebra

3.1

Steps 1.2 and 1.4, read through step 2.1, show that centering and scaling the nodes improves the conditioning of the same interpolation problem by four orders of magnitude.

step 2.1∎

Depends on

Used by

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Dependency tree · two levels

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Sources

L. N. Trefethen and D. Bau III, Numerical Linear Algebra, Lecture 18 (standard reference, not scraped)